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The washer method

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  • 1. Solids of Revolution: Volume Review: The Disk Method 1
  • 2. The Disk Method If a region in the plane is revolved about a line, the resulting solid is a solid of revolution, and the line is called the axis of revolution. The simplest such solid is a right circular cylinder or disk, which is formed by revolving a rectangle about an axis adjacent to one side of the rectangle, as shown in Figure 7.13. Figure 7.13 2
  • 3. The Disk Method Divide the region into n equal disks Find the volume of each disk, to get an approximate volume. As the number of disks approaches infinity, we approach the actual volume: ANIMATION Figure 7.13 3
  • 4. The Washer Method 4
  • 5. The Washer Method The disk method can be extended to cover solids of revolution with holes by replacing the representative disk with a representative washer. The washer is formed by revolving a rectangle about an axis, as shown in Figure 7.18. If r and R are the inner and outer radii of the washer and w is the width of the washer, the volume is given by Volume of washer = π(R2 – r2)w. 5 Figure 7.18
  • 6. The Washer Method To see how this concept can be used to find the volume of a solid of revolution, consider a region bounded by an outer radius R(x) and an inner radius r(x), as shown in Figure 7.19. Figure 7.19 6
  • 7. Example 3 – Using the Washer Method Find the volume of the solid formed by revolving the region bounded by the graphs of about the x-axis, as shown in Figure 7.20. Figure 7.20 7
  • 8. The Washer Method If the region is revolved about its axis of revolution, the volume of the resulting solid is given by Note that the integral involving the inner radius represents the volume of the hole and is subtracted from the integral involving the outer radius. 8
  • 9. Example 3 – Solution In Figure 7.20, you can see that the outer and inner radii are as follows. Integrating between 0 and 1 produces 9
  • 10. Example 3 – Solution cont’d 10
  • 11. Practice Find the volume of the solid formed by revolving the region bounded by the graphs of f(x)= - x^2 +5x+3 and g(x) = -x + 8 about the x-axis. SETUP the integral and then finish with your calculator. Approximately 442.34 cubic units 11
  • 12. The Washer Method So far, the axis of revolution has been horizontal and you have integrated with respect to x. In the Example 4, the axis of revolution is vertical and you integrate with respect to y. In this example, you need two separate integrals to compute the volume. 12
  • 13. Example 4 – Integrating with Respect to y, Two-Integral Case Find the volume of the solid formed by revolving the region bounded by the graphs of y = x2 + 1, y = 0, x = 0, and x = 1 about y-axis, as shown in Figure 7.21. Figure 7.21 13
  • 14. Example 4 – Solution For the region shown in Figure 7.21, the outer radius is simply R = 1. There is, however, no convenient formula that represents the inner radius. When 0 ≤ y ≤ 1, r = 0, but when 1 ≤ y ≤ 2, r is determined by the equation y = x2 + 1, which implies that 14
  • 15. Example 4 – Solution cont’d Using this definition of the inner radius, you can use two integrals to find the volume. 15
  • 16. Example 4 – Solution Note that the first integral cont’d represents the volume of a right circular cylinder of radius 1 and height 1. This portion of the volume could have been determined without using calculus. 16
  • 17. Solids with Known Cross Sections 17
  • 18. Solids with Known Cross Sections With the disk method, you can find the volume of a solid having a circular cross section whose area is A = πR2. This method can be generalized to solids of any shape, as long as you know a formula for the area of an arbitrary cross section. Some common cross sections are squares, rectangles, triangles, semicircles, and trapezoids. 18
  • 19. Solids with Known Cross Sections Figure 7.24 19
  • 20. Example 6 – Triangular Cross Sections Find the volume of the solid shown in Figure 7.25. The base of the solid is the region bounded by the lines and x = 0. Figure 7.25 The cross sections perpendicular to the x-axis are equilateral triangles. 20
  • 21. Example 6 – Solution The base and area of each triangular cross section are as follows. 21
  • 22. Example 6 – Solution cont’d Because x ranges from 0 to 2, the volume of the solid is 22